What you're looking for is called Zipf's law. This law says that many distribution curves in which the data values are placed in rank order on the horizontal axis by frequency (or, equivalently, percent) follow a power law. The most famous use of Zipf's law is to describe the frequency of word usage in any given language, although the Wikipedia article specifically mentions income rankings as you ask for.
It can be thought of as a discrete version of the Pareto distribution, so you're right about that. Added: This is because Zipf's law is the discrete power law distribution, and Pareto is the continuous power law distribution.
(Update, in response to the OP's request for more on the relationship between Zipf and Pareto.)
I'm going to do this in the general case. The argument will also be for numbers and amounts, rather than probabilities, with the understanding that we can convert the functions involved to pdfs or pmfs by scaling by the appropriate constants.
Suppose we have the density function $p(x)$ for dollars (although it could be any good) allocated among people in a group, so that $\int_a^b p(x) dx$ gives the number of people in the group who have between $a$ and $b$ dollars. Now, rank the people in the group by wealth, and let $z(y)$ denote the wealth that the person ranked $y$ has. The question then is, "What is the relationship between $p(x)$ and $z(y)$?"
Consider the number of people who have more than $M$ dollars. Using $p(x)$, that is given by $\int_M^{\infty} p(x) dx$. But this is also $R$, where $R$ is the largest rank of a person who has at least $M$ dollars. (In other words, if the 34th person has at least $M$ but the 35th does not, then $R = 34$.) So $z(R+1) < M \leq z(R)$. If the population is large enough, we can say $z(R) = M$ without losing much accuracy. Thus $R = z^{-1}(M)$. So there's our relationship (approximately): $$\int_M^{\infty} p(x) dx = z^{-1}(M).$$
Thus the ranking function $z(y)$ is the inverse of the wealth tail cumulative distribution function $\int_M^{\infty} p(x) dx$.
How does this relate to power laws? Well, in this special case, if $p(x) = \frac{C}{x^{\alpha+1}}$ (i.e., a Pareto distribution) for some $\alpha > 0$ and constant $C$, then we have $$z^{-1}(M) = \int_M^{\infty} \frac{C}{x^{\alpha}} dx = \frac{C}{\alpha M^{\alpha}},$$
which means, for some constant $K$, $$z(y) = \frac{K}{y^{\frac{1}{\alpha}}}.$$
Thus $z(y)$ is also a power law. Thus a Pareto (power law) distribution function for some good produces a power law ranking function for people with that good (i.e., Zipf).
A discrete Pareto distribution is normally known as the Riemann Zeta distribution which is ill-suited to your problem because it is unbounded above … whereas you seek a finite upper bound of $n=10000$.
Your other suggestion of a Zipf distribution seems more appropriate as it has a finite upper bound at $n$ elements, so let us use that. I am using the mathStatica/Mathematica combo here …
So, let $X \sim \text{Zipf}(n,a)$ with pmf $f(x)$:
f = 1/(HarmonicNumber[n, a] x^a);
domain[f] = {x, 1, n} && {n>0, a>0} && {Discrete};
The cdf, $P(X \leq x)$, is:
Prob[f]
You wish to find the value of parameter 'a' such that $P(X \leq 2000) = 0.8$, given $n=10000$. Solving this numerically yields:
So that's it … a $\text{Zipf}(n,a)$ pmf with $n = 10000$ and $a = 0.949588$, which will yield you:
$P(X \leq 10000) = 1$
$P(X \leq 2000) = 0.8$
Estoup
Finally, depending on how important it is that $P(X \leq 2000) = 0.8$, you might want to consider the simplified case of $a = 1$ which is known as an Estoup distribution with pmf:
f = 1/(HarmonicNumber[n] x)
and which will yield:
I'm struggling to plot that cdf in wolfram mathematica... do you know the syntax for that?
Sure. Here is the cdf in InputForm ... just copy and paste:
F = Piecewise[{{1, x >= n}, {(-HurwitzZeta[a, 1 + Floor[x]] + Zeta[a])/HarmonicNumber[n, a], 1 <= x < n}}, 0]
If $n$ is small, you can see that this has a stepped structure:
Plot[F /. {n -> 14, a -> 2}, {x, 0, 14}, PlotRange -> All]
And, for your case, with very large n, it appears continuous, but is of course actually discrete, if you were to zoom in:
Plot[F /. {n -> 10000, a -> 0.949588}, {x, -10, 11000}, PlotRange -> All]
Best Answer
In light of the fact that $$\frac{1}{x^\alpha}\to 0\text{ as }x\to\infty\quad\text{ provided } \alpha>0$$
, we have the $r$-th order raw moment of $X$ about $0$ :
\begin{align} E(X^r)&=\int_{\theta}^\infty \frac{x^r\,k\theta^k}{x^{k+1}}\,dx \\&=k\theta^k\int_{\theta}^\infty x^{r-k-1}\,dx \\&=k\theta^k\lim_{A\to\infty}\left[\frac{x^{-(k-r)}}{-(k-r)}\right]_{\theta}^A \\&=\frac{k\theta^r}{k-r}\qquad,\text{ if }k>r \end{align}
You should be able to proceed now.